# Finding Phase angle of Simple Harmonic Motion?

A sinusoidal oscillator has :

$$x=x_{max} \cos(\omega t - \varphi )$$

Period is 2, initial displacement is 100mm initial velocity is 200mm/s

What is the phase angle assuming $-\pi < \varphi < \pi$

How do I go about solving this?

Is the phase $(\omega t - \varphi)$? But I do not know what $x_{max}$ is, how am I supposed to solve for the angle?

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$x_{max}$ is the amplitude of the oscillations, and yes, ${\omega}t - \varphi$ is the phase.

We know that the period $T$, is the reciprocal of the frequency $f$, or $$T = 1/f$$

We also know that $\omega$, the angular frequency, is equal to $2\pi$ times the frequency, or $$\omega = 2{\pi}f$$

From here, we can use the initial conditions to find the amplitude.

$x(0) = x_{max}cos(\varphi)$

$\dot{x}(0) = {\omega}x_{max}sin(\varphi)$

From here it should be a simple matter to find $\varphi$.

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but what is x_max –  Fendi Feb 19 '12 at 1:11
@Fendi, try using the advice Daniel gave you. If you can't figure out how to work out the problem given that, then you can come back and ask for clarification. –  David Z Feb 19 '12 at 2:21
Thats exactly why I posted back asking for clarification. I didn't get it. –  Fendi Feb 19 '12 at 3:04
I've specified what $x_{max}$ is at the beginning of the post; it is the amplitude of the oscillations, i.e. the maximum displacement of the particle from its 0 position. If you mean how do you find the value of it; you use the initial conditions specified to find both $x_{max}$ and $\varphi$, the same way you'd find the values of any two equations with two unknowns. –  Daniel Blay Feb 19 '12 at 7:50
OFCOURSE !!! Simultaneous equations !!! Can you please clarify the following : Isn't the equation supposed to be x(0)=x_max cos (-phi) ? where phi is negative (wt - phi) ? and how do you get a w when you derive x(0)=x_max cos (-phi) ? is the derivative of phi the angular frequency ? –  Fendi Feb 19 '12 at 15:01